Say you have an array for which the ith element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
- 题意为低价买入高价卖出,求最大收益
- 卖出必须在买入之后
- 我们要理解一个问题,比如数组为[9 8 7 6 1 2 3 4]
那么我们从1买入,从4卖出是符合条件的最大收益
同时我们可以1买入2卖出,2买入3卖出,3买入4卖出,收益还是一样的 - 因此我们得出可以遍历一遍后项比前项大的差累加就好了
class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.size()<=1)
return 0;
int pmax = 0;
for(int i = 1;i<prices.size();i++) {
if(prices[i]>prices[i-1]) {
pmax = pmax + (prices[i]-prices[i-1]);
}
}
return pmax;
}
};
本文介绍了一种计算股票交易最大利润的算法。该算法允许进行多次买卖操作,但每次卖出后才能再次购买。通过一次遍历数组,比较相邻元素大小并累加差值来获得最大利润。
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